How to derive the noninformative prior for location parameters and scale parameter? I am reading this paper, in it:

I have a lot of confusion reading it, I will list it one by one:


*

*Let $X$ be distributed as $f(x-\theta)$, which is a location invariant density. 



Q1: The sentence which is a location invariant density, is this a
  assumption or just a claim? If this is a claim, then Gaussian can be
  written in this form but apparently it is a location dependent
  distribution.)



*Since the model is location invariant, the prior distribution should be location invariant.



Q2: why so easily the conclusion is draw: the prior distribution
  should be location invariant because the model is location
  invariant. Why bother introduce $Y$?



Q3: why This lead to $\pi(\sigma)=\frac{1}{c}\pi(\frac{\sigma}{c})$
  but not just $\pi(\sigma)=\pi(\frac{\sigma}{c})$? How it is derived, the above formula has no dependence on $\sigma$ but the below one suddenly has the dependence on it? If I just let $A=\sigma$, then it should be $\pi(\sigma)=\pi(\frac{\sigma}{c})$.

 A: Regarding Q1, I think of "invariant" as meaning that the density doesn't change if both the observation and parameter are changed in a "compatible" way. Going back to your example of a Gaussian distribution with a known variance and unknown mean, if you shift both the observation and mean parameter by adding some constant number to both of them, you'll end up with the same density height. 
It's just an assumption of a model we're working with. Sometimes, when we are working under the Gaussian assumption, it is useful to assume that the variance parameter is not known.
Q2 is a little trickier. If the likelihood is location invariant, that means it only gives us information regarding how far away the observation is from the center. If we want to use a noninformative prior, it should not give us information about, say, the scale. If it did, then that information would remain "unchecked" after we use Bayes' theorem to go from the prior to the posterior. If it does, then this doesn't seem like the prior is noninformative.
Q3 has to do with the transformation theorem. Keep in mind the difference between the argument to the density function, and the random variable that the density describes. If we have a prior for the random variable $\sigma$, and we plug in argument $\sigma$, we evaluate $\pi(\sigma)$. If we transform $\sigma \to \theta = c\sigma$ for some positive constant $c$, then the density for $\theta$ can be written in terms of the original density as
$$
\pi_{\theta}(t) = \pi_{\sigma}(t/c) \frac{1}{c}.
$$
To verify their claim that the prior should be proportional to $1/c$, just plug some argument into both of the priors, and check that they give you the same density height:
\begin{align*}
\pi_{\theta}(a) &= \pi_{\sigma}(a)  \\
\pi_{\sigma}(a/c) \frac{1}{c} &= \pi_{\sigma}(a) \\
\frac{1}{a/c} \frac{1}{c} &= \frac{1}{a} .
\end{align*}
A: The first two questions were answered by Taylor (using natural language), here I only give another answer for Q3, using math formulas.
To reflect scale invariance of density of $X$:
$$p(x|\sigma)=\frac{1}{\sigma}f\left(\frac{x}{\sigma}\right),$$
the prior density of $\sigma$ should assign equal probability mass to an interval $A\le\sigma\le B$ as to a scaled interval $A/c\le\sigma\le B/c,\ c>0$. Thus we have
$$\int_A^B\pi(\sigma)d\sigma=\int_{A/c}^{B/c}\pi(\sigma)d\sigma=\int_A^B\frac{1}{c}\pi\left(\frac{\sigma}{c}\right)d\sigma,$$
where the last equality is change of variable $\sigma'=c\sigma$ in integral and changing the notation of new variable $\sigma'$ back to $\sigma$.
Because this equaltion must hold for any choices of $A$ and $B$, we have
$$\pi(\sigma)=\frac{1}{c}\pi\left(\frac{\sigma}{c}\right), \ c>0$$
due to fundamental theorem of calculus. This is the first part of what is led to in the excerpted paper.
The following derives why the above identity implies $\pi(\sigma)=\sigma^{-1}$. By definition of derivative, we have
$$\pi'(\sigma)=\lim\limits_{\Delta\sigma\to0}\frac{\pi(\sigma+\Delta\sigma)-\pi(\sigma)}{\Delta\sigma}.$$
But,
$$\begin{align*}
\pi(\sigma+\Delta\sigma)&=\pi\left(\bigl(1+\frac{\Delta\sigma}{\sigma}\bigr)\sigma\right)\\
&=\pi\left(\frac{\sigma}{\frac{1}{1+\frac{\Delta\sigma}{\sigma}}}\right)\\
&=\frac{1}{1+\frac{\Delta\sigma}{\sigma}}\pi(\sigma)\\
&=\frac{\sigma}{\sigma+\Delta\sigma}\pi(\sigma),
\end{align*}$$
where we used $\pi\left(\frac{\sigma}{c}\right)=c\pi(\sigma)$ which we have proved and in which $c=\frac{1}{1+\frac{\Delta\sigma}{\sigma}}$.
Plugging the expression for $\pi(\sigma+\Delta\sigma)$ into the definition of $\pi'(\sigma)$, we have
$$\begin{align*}
\frac{\pi(\sigma+\Delta\sigma)-\pi(\sigma)}{\Delta\sigma}&=\frac{\left(\frac{\sigma}{\sigma+\Delta\sigma}-1\right)\pi(\sigma)}{\Delta\sigma}\\
&=-\frac{\pi(\sigma)}{\sigma+\Delta\sigma}.
\end{align*}$$
Taking limit $\Delta\sigma\to0$ on both sides, we have
$$\pi'(\sigma)=-\frac{\pi(\sigma)}{\sigma},$$
or $$\frac{\pi'(\sigma)}{\pi(\sigma)}=-\frac{1}{\sigma}.$$
This simple ODE can be easily solved that $\pi(\sigma)=\frac{1}{\sigma}$ if we ignore the constant, which is the second part of what is led to in the excerpted paper.
